Question: Consider the polar curve $r=1+2\sin(\theta)$. What is the equation of the tangent line to the curve $r$ at $\theta = \dfrac{5\pi}{6}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-1=3\sqrt{3}\big(x-\sqrt{3}\big)$ (Choice B) B $y-1=-3\sqrt{3}\big(x+\sqrt{3}\big)$ (Choice C) C $y-1=\sqrt{3}\big(x-\sqrt{3}\big)$ (Choice D) D $y-1=-\sqrt{3}\big(x+\sqrt{3}\big)$
Answer: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Then we can use the point-slope form to complete the equation for the tangent line through $\theta=\dfrac{5\pi}{6}$. For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={(1+2\sin\theta)}\cos(\theta) \\\\ y&={(1+2\sin\theta)} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{2\sin(2\theta)+\cos(\theta)}{2\cos(2\theta)-\sin(\theta)} \end{aligned}$ Evaluating $\dfrac{dy}{dx}$ at ${\theta = \dfrac{5\pi}{6}}$ gives us the slope of our tangent line. $\begin{aligned} {\left. \dfrac{dy}{dx}\right| _{\theta=\tfrac{5\pi}{6}}}&=\dfrac{2\sin\left(2\left({\dfrac{5\pi}{6}}\right)\right)+\cos\left({\dfrac{5\pi}{6}}\right)}{2\cos\left(2\left({\dfrac{5\pi}{6}}\right)\right)-\sin\left({\dfrac{5\pi}{6}}\right)} \\\\ &=\dfrac{2\sin\left(\dfrac{5\pi}{3}\right)+\cos\left({\dfrac{5\pi}{6}}\right)}{2\cos\left(\dfrac{5\pi}{3}\right)-\sin\left({\dfrac{5\pi}{6}}\right)} \\\\ &=\dfrac{2\left(-\dfrac{\sqrt{3}}{2}\right)-\dfrac{\sqrt{3}}{2}}{2\left(\dfrac{1}{2}\right)-\dfrac{1}{2}} \\\\ &={-3\sqrt{3}} \end{aligned}$ We now find $x$ and $y$ at the point $\theta=\dfrac{5\pi}{6}$. $\begin{aligned} {x\left({\dfrac{5\pi}{6}}\right)}&=\left(1+2\sin\left({\dfrac{5\pi}{6}}\right)\right)\cos\left({\dfrac{5\pi}{6}}\right) \\\\&=\left(1+2\left(\dfrac{1}{2}\right)\right)\left(-\dfrac{\sqrt{3}}{2}\right) \\\\ &={-\sqrt{3}} \\\\ \\\\ y\left({\dfrac{5\pi}{6}}\right)}&=\left(1+2\sin\left({\dfrac{5\pi}{6}}\right)\right) \sin\left({\dfrac{5\pi}{6}}\right) \\\\ &=\left(1+2\left(\dfrac{1}{2}\right)\right)\left(\dfrac{1}{2}\right) \\\\ &=1} \end{aligned}$ Therefore the equation of our tangent line is: $\begin{aligned} y-1}&={-3\sqrt{3}}\left(x-({-\sqrt{3}})\right) \\\\ y-1&=-3\sqrt{3}\left(x+\sqrt{3}\right) \end{aligned}$ The graph of the tangent is shown.